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The definition in my text reads,

An integer $d$ is said to be the greatest common divisor of two non-zero integers $a$ and $b$ iff,

$d|a$ and $d|b$ and if $k$ is any other common divisor of $a$ and $b$ then $k|d$

Now here's the thing, if $d|a$ and $d|b$ then surely $-d|a$ and $-d|b$ as well, also $k|-d$

What I take from this? GCD is not unique! That is if $\mathrm{gcd}(12,8)= 4$ then by the definition, $\mathrm{gcd}(12,8) = -4$ as well.

Yet I never ever seen a negative gcd. Someone please explain.

Maybe, $4>-4$, and we want the "greatest common factor" so...? But that still doesn't justify the definition.

Jyerki menon
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William
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    Where is than definition from? From wikipedia: "In mathematics, the greatest common divisor (gcd) of two or more integers, which are not all zero, is the largest positive integer that divides each of the integers" – Roberto Rastapopoulos Oct 02 '18 at 10:15
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    I expect your definition required $d\in \mathbb N$, – lulu Oct 02 '18 at 10:18
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    Indeed, the greatest common divisor of two elements is only uniquely determined up to an invertible element (i.e. up to a sign in the integers). For the integers we usually choose the positive one, since this is a "nice" way to single out one of them. – Tobias Kildetoft Oct 02 '18 at 10:20
  • @RobertoRastapopoulos you can find it many algebra textbooks, but here is an example in the form of question. https://math.stackexchange.com/q/85565/552998 – William Oct 02 '18 at 10:24
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    In some definitions it's a gcd (because there can be more than one) – John Cataldo Oct 02 '18 at 10:42
  • @TobiasKildetoft But then again, if you agree that gcd can be positive as well as negative, obviously, positive > negative. So the positive divisor should be considered as it's greater. How can gcd be negative in the first place, as it asks for the "greatest" factor. – William Oct 02 '18 at 12:32
  • The text forgot to say that $d$ is supposed to be a nonnegative integer. It is also imprecise, because the assumption that $a$ and $b$ are nonzero is not required. – egreg Oct 03 '18 at 22:45

2 Answers2

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The text is employing the $\color{#c00}{\rm universal}$ definition of a gcd, namely

$$\ c\mid a,b \!\!\color{#c00}{\overset{\rm u\!\!}\iff}\! c\mid \gcd(a,b)\qquad$$

Direction $(\Leftarrow)$ implies a gcd is a common divisor of $a,b,\,$ by choosing $ c = \gcd(a,b),\,$ and the reverse direction $(\Rightarrow)$ implies that a gcd is "greatest" w.r.t. divisibility order, i.e. it's divisible by all other common divisors $\,c\,$ of $a,b\,$ (so has greatest magnitude in $\Bbb Z,$ and greatest degree in $K[x])$.

Generally a gcd is not unique: if $\,d,d'$ are both gcds of $\,a,b\,$ then $\, c\mid d\!\!\!\color{#c00}{\overset{\rm u\!\!}\iff}\! c\mid a,b\!\!\!\color{#c00}{\overset{\rm u\!\!}\iff}\! c\mid d'\,$ so specializing $\,c =d\,$ and $\,c = d'\,$ shows $\,d\mid d'\mid d,\,$ i.e. $\,d\sim d'\,$ are associate (divide each other). And conversely too: $\,d=\gcd(a,b)\,$ is associate to $\, d'\,$ then $\,d\mid d'\mid d,\,$ so $\,c\mid d\!\iff\! c\mid d',\,$ thus $\,d'$ is also a gcd of $\,a,b.\,$ In an integral domain $\,a\,$ is associate to $\,b\!\iff\!$ $\,a = ub\,$ where $\,u\,$ is a unit (= invertible), i.e. associates are unit multiples. Thus gcds are unique up to unit multiples.

In some rings with simple unit group structure we can choose canonical representatives of associate classes, which allows is to choose normal forms for gcds, e.g. in $\,\Bbb Z\,$ (with units $\pm 1)$ we normalize gcds $\ge 0,\,$ and in a polynomial ring $\,K[x]\,$ over a field (units = constants $0\neq c\in K) $ we normalize polynomial gcds to be monic (lead coeff $\,c_n = 1),\,$ by scaling the polynomial by $\,c_n^{-1}\,$ if need be (so a constant gcd $\,c_0\neq 0\,$ normalizes to $1).\,$ Hence in both cases we can say that two elements are coprime $\!\iff\!$ their gcd $= 1$ (vs. a unit). Such normalizations are sometimes called unit normal representatives in the literature.

Note on notation (abuse of language). Some writers (implicitly) use the convention that $\gcd(a,b) = c\,$ means $\,\gcd(a,b)\approx c,\,$ i.e. the gcd is associate to $c,\,$ even when there is no natural choice of unit normalization. This is similar to other such abuses of language that - for convenience - blur the distinction between an equivalence class and a (non-canonical) rep of the class. Common simple examples are fractions: $\,x = a/b,\,$ meaning $\,x\approx a/b\,$ or $\, x = [a/b],\,$ and elements of quotient rings, e.g. in $\,R = \Bbb Z/n\,$ one often denotes the coset $[k] = k+n\Bbb Z\,$ by $\,k\,$ and $\,j = k\,$ means $\,[j] = [k]\,$ (or $\,j\equiv_n k\,$ when using congruences), and such notational abuse is often (naturally) extended to $R$-algebras over such quotient rings, e.g. polynomial and power series rings, where we often see $\,ax^2+bx+c\,$ vs. $[a] x^2 + [b]x + [c],\,$ and similarly for product and matrix rings (tuple notation), etc. Without such, notational complexity might serve to obfuscate the key idea(s) under discussion.

Bill Dubuque
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You are right, it is all about definitions. With the definition you gave the gcd is really not unique and it might be negative as well. But because most of the time we are using only the positive gcd then some simply prefer to add the words "$d$ is positive" to your definition or just give other definitions. For example a very common definition of gcd in number theory is "$d|a$ and $d|b$ and if $k$ is any other common divisor of $a$ and $b$ then $k\leq d$". So that definition already requires the gcd to be positive.

Mark
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  • But then again, if you agree that gcd can be positive as well as negative, obviously, positive > negative. So the positive divisor should be considered as it's greater. How can gcd be negative in the first place, as it asks for the "greatest" factor. – William Oct 02 '18 at 12:31
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    So again, it's up to how you define. The definition I gave you obviously requires the gcd to be positive and hence unique. But in other places you might find a definition that doesn't require that and allows the gcd to be negative as well. (your definition for example). So all the question is do you call the negative number a gcd. Some people do, some do not. The fact is we usually don't really need the negative gcd so in most books you will see a definition according to which the gcd must be positive. – Mark Oct 02 '18 at 12:37
  • Thank you :) I'm just having a hard time accepting that a particular concept has 2 definition, both of them are slightly different from one another. This just doesn't seem math-likely though... – William Oct 02 '18 at 14:55
  • Well, note that all the definitions of the positive gcd are equivalent, there are no contradictions here. It's just that question-do you want to call the negative number a gcd or not? It's all just about names. I prefer the gcd to be positive and unique, it's much easier to think this way. – Mark Oct 02 '18 at 16:40
  • @William: start to get used to this. If you go beyond the reign of delle established math, there will be things with different definitions, maybe given by different people, and sometimes they coincide only in special cases. The thing is mathematicians look at the concept behind, and formalizations are just an attempt the capture its main features. Sometimes different definitions capture different features of the same concept. – Andrea Marino May 06 '21 at 07:04